chemistry
posted by Anonymous .
A solution contains 10.05 grams of an unknown covalent substance in 50.0
grams of water. The freezing point of this solution is () 3.16o C.
What is the formula weight of the unknown compound?
Below is what I tried, but don't know what to do next. Please explain stepbystep. Thanks
m=3.16/1.86=1.69 moles of solute/kg of water. 10.05 x 1000*50 g/kg of water is 200.4 g. So in l liter we have 1.69 moles = 200.4g. Mass of 1 mole =200.4/1.69 = 118.59

I don't follow all of you steps but you have essentially the right answer.
10.05 x 1000*50g/kg is not 200.4 and and kg is 50/1000 and not 50*1000. Anyway, here is the way to do it right.
dT = Kf*m
3.16/1.86 = m = 1.6989 (you threw the last 89 away and you should have rounded up to 1.70).
Tnen m = mols/kg solvent or
m x kg solvent = mols = 1.70 x 0.05 kg = 0.0849 mols.
Then mols = grams/molar mass or
molar mass = grams/mols = 10.05/0.0849 = 118.4.
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